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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

• (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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Example Physics Problems and Solutions

Learning how to solve physics problems is a big part of learning physics. Here’s a collection of example physics problems and solutions to help you tackle problems sets and understand concepts and how to work with formulas:

Physics Homework Tips Physics homework can be challenging! Get tips to help make the task a little easier.

Unit Conversion Examples

There are now too many unit conversion examples to list in this space. This Unit Conversion Examples page is a more comprehensive list of worked example problems.

Newton’s Equations of Motion Example Problems

Equations of Motion – Constant Acceleration Example This equations of motion example problem consist of a sliding block under constant acceleration. It uses the equations of motion to calculate the position and velocity of a given time and the time and position of a given velocity.

Equations of Motion Example Problem – Constant Acceleration This example problem uses the equations of motion for constant acceleration to find the position, velocity, and acceleration of a breaking vehicle.

Equations of Motion Example Problem – Interception

This example problem uses the equations of motion for constant acceleration to calculate the time needed for one vehicle to intercept another vehicle moving at a constant velocity.

Vertical Motion Example Problem – Coin Toss Here’s an example applying the equations of motion under constant acceleration to determine the maximum height, velocity and time of flight for a coin flipped into a well. This problem could be modified to solve any object tossed vertically or dropped off a tall building or any height. This type of problem is a common equation of motion homework problem.

Projectile Motion Example Problem This example problem shows how to find different variables associated with parabolic projectile motion.

Accelerometer and Inertia Example Problem Accelerometers are devices to measure or detect acceleration by measuring the changes that occur as a system experiences an acceleration. This example problem uses one of the simplest forms of an accelerometer, a weight hanging from a stiff rod or wire. As the system accelerates, the hanging weight is deflected from its rest position. This example derives the relationship between that angle, the acceleration and the acceleration due to gravity. It then calculates the acceleration due to gravity of an unknown planet.

Weight In An Elevator Have you ever wondered why you feel slightly heavier in an elevator when it begins to move up? Or why you feel lighter when the elevator begins to move down? This example problem explains how to find your weight in an accelerating elevator and how to find the acceleration of an elevator using your weight on a scale.

Equilibrium Example Problem This example problem shows how to determine the different forces in a system at equilibrium. The system is a block suspended from a rope attached to two other ropes.

Equilibrium Example Problem – Balance This example problem highlights the basics of finding the forces acting on a system in mechanical equilibrium.

Force of Gravity Example This physics problem and solution shows how to apply Newton’s equation to calculate the gravitational force between the Earth and the Moon.

Coupled Systems Example Problems

Coupled systems are two or more separate systems connected together. The best way to solve these types of problems is to treat each system separately and then find common variables between them. Atwood Machine The Atwood Machine is a coupled system of two weights sharing a connecting string over a pulley. This example problem shows how to find the acceleration of an Atwood system and the tension in the connecting string. Coupled Blocks – Inertia Example This example problem is similar to the Atwood machine except one block is resting on a frictionless surface perpendicular to the other block. This block is hanging over the edge and pulling down on the coupled string. The problem shows how to calculate the acceleration of the blocks and the tension in the connecting string.

Friction Example Problems

These example physics problems explain how to calculate the different coefficients of friction.

Friction Example Problem – Block Resting on a Surface Friction Example Problem – Coefficient of Static Friction Friction Example Problem – Coefficient of Kinetic Friction Friction and Inertia Example Problem

Momentum and Collisions Example Problems

These example problems show how to calculate the momentum of moving masses.

Momentum and Impulse Example Finds the momentum before and after a force acts on a body and determine the impulse of the force.

Elastic Collision Example Shows how to find the velocities of two masses after an elastic collision.

It Can Be Shown – Elastic Collision Math Steps Shows the math to find the equations expressing the final velocities of two masses in terms of their initial velocities.

Simple Pendulum Example Problems

These example problems show how to use the period of a pendulum to find related information.

Find the Period of a Simple Pendulum Find the period if you know the length of a pendulum and the acceleration due to gravity.

Find the Length of a Simple Pendulum Find the length of the pendulum when the period and acceleration due to gravity is known.

Find the Acceleration due to Gravity Using A Pendulum Find ‘g’ on different planets by timing the period of a known pendulum length.

Harmonic Motion and Waves Example Problems

These example problems all involve simple harmonic motion and wave mechanics.

Energy and Wavelength Example This example shows how to determine the energy of a photon of a known wavelength.

Hooke’s Law Example Problem An example problem involving the restoring force of a spring.

Wavelength and Frequency Calculations See how to calculate wavelength if you know frequency and vice versa, for light, sound, or other waves.

Heat and Energy Example Problems

Heat of Fusion Example Problem Two example problems using the heat of fusion to calculate the energy required for a phase change.

Specific Heat Example Problem This is actually 3 similar example problems using the specific heat equation to calculate heat, specific heat, and temperature of a system.

Heat of Vaporization Example Problems Two example problems using or finding the heat of vaporization.

Ice to Steam Example Problem Classic problem melting cold ice to make hot steam. This problem brings all three of the previous example problems into one problem to calculate heat changes over phase changes.

Charge and Coulomb Force Example Problems

Electrical charges generate a coulomb force between themselves proportional to the magnitude of the charges and inversely proportional to the distance between them. Coulomb’s Law Example This example problem shows how to use Coulomb’s Law equation to find the charges necessary to produce a known repulsive force over a set distance. Coulomb Force Example This Coulomb force example shows how to find the number of electrons transferred between two bodies to generate a set amount of force over a short distance.

Gurumuda Networks

Newton’s first law of motion – problems and solutions

1. A person is in an elevator that moving upward at a constant velocity . The weight of the person is 800 N. Immediately the elevator rope is broke, so the elevator falls. Determine the normal force acted by elevator’s floor to the person just before and after the elevator’s rope broke.

A. 800 N and 0

B. 800 N and 800 N

C. 1600 N and 0

D. 1600 N and 800 N

Weight (w) = 800 Newton

Wanted: The normal force (N)

Before the elevator’s rope broke

When the person stands on the floor of the elevator, weight acts on the person where the direction of the person is downward. That person at rest so that there must a normal force acts on the person, where the direction of the normal force is upward and the magnitude of the normal force same as the magnitude of the weight.

N = 800 Newton

After the elevator’s rope broke

After the elevator’s rope broke, the elevator and the person free fall together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.

The correct answer is A.

2. A block with a mass of 20 gram moves at a constant velocity on a rough horizontal floor at a constant velocity if there is an external force of 2 N acts on the block. Determine the magnitude of the friction force experienced by the block.

Mass (m) = 20 gram

Force (F) = 2 Newton

Wanted: Magnitude of friction force experienced by the block.

– The magnitude of friction force (F fric ) same as the magnitude of the external force (F)

– The friction force (F fric ) has opposite direction with the external force (F)

Apply Newton’s first law of motion :

F – F fric = 0

F fric = 2 Newton

The correct answer is C.

3. A smooth inclined plane with the length of 0.6 m and height of 0.4 m. A block with the weight of, 1350 N will move upward using the inclined plane. Determine the magnitude of force need to move the block.

Weight of block (w) = 1350 Newton

hyp = 0.6 m

opp = 0.4 m

Wanted : The minimum force

opp = bc = 0.4 m

Sin θ = bc / ac = 0.4 / 0.6 = 4/6 = 2/3

Based on Newton’s first law of motion, the block start to moves upward then the external force (F) minimal same as the horizontal component of weight (w x ).

F – w x = 0

If F = w x , then object start to moving upward at constant velocity.

w x = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton

The correct answer is D.

4. Three forces, F 1 = 22 N, F 2 = 18 N and F 3 = 40 N act on a block. Which figure describes Newton’s first law.

Newton’s first law : Net force (ΣF) = 0.

A. F 1 + F 2 – F 3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0

B. F 2 + F 3 – F 1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

C. F 2 + F 3 – F 1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)

D. F 1 + F 3 – F 2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)

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